On 05/28/2011 01:16 PM, Steven Watanabe wrote:
AMDG
On 05/28/2011 12:42 PM, Pyry Jahkola wrote:
On 2011-05-28 21:59:57 +0300, Matt Calabrese said:
On Sat, May 28, 2011 at 2:55 PM, Steven Watanabe <watanabesj@gmail.com>wrote:
I don't have the concepts proposal handy, but doesn't LessThanComparable have an axiom that requires that (a< b) is equivalent to (b> a)?
Yes.
axiom Consistency(T a, T b) { (a> b) == (b< a); (a<= b) == !(b< a); (a>= b) == !(a< b); }
For what it's worth, e.g. NaN's are an exception to that rule...
It's not an exception. The axiom only applies to LessThanComparable types. Technically, double isn't LessThanComparable, because it violates this rule.
Given that double is a commonly used type (and the fact that it has an order is also often important), this does suggest that we may not want to always guarantee that these consistency conditions hold, or may want to allow certain values such as NaN to not satisfy the consistency conditions while still allowing using the of type.